Let $\{a_1,...,a_n\}$ be commuting normal operators on a Hilbert space. Put $A:= C^*(1,a_1,...,a_n)$. By Gelfand theorem ,abelian C*-algebra $A$ is identified with the algebra $C(\Omega)$ of all continuous functions on spectrum $\Omega$ of $A$, which is compact. Consider the map $\Gamma : \Omega \to \Bbb C ^n$ such that $\Gamma(\phi):=(\phi(a_1),...,\phi(a_n))$. Clearly we can identify $\Omega$ and $\Gamma(\Omega)$. Now I claim $\Gamma(\Omega) = \sigma(a_1)\times ... \times \sigma(a_n)$.

To show it we have $$\phi(a_i) - a_i = a_i\phi(1) - a_i =0$$ for every $i=1,...,n$, and $\phi\in\Omega$. So clearly $\Gamma(\Omega)\subset \sigma(a_1)\times ... \times \sigma(a_n)$.

Conversely, Let $(\lambda_1,...,\lambda_n)\in \sigma(a_1)\times ... \times \sigma(a_n)$. Define $\phi: A\to \Bbb C$ such that $\phi(a_i):=\lambda_i$ for $i=1,...,n$ and extend it to a $*-$ homomorphism. Now could I claim $\phi$ is a character on $A$?

Thanks in advance.


This is not true in general, e.g. let $a_1=\begin{pmatrix}\sqrt{2}/2&-\sqrt{2}/2\\\sqrt{2}/2&\sqrt{2}/2\end{pmatrix}$ and $a_2=a_1^*$ in $M_2(\mathbb{C})$. Since $a_1$ is unitary, then $A=C^*(1,a_1,a_2)=C^*(a_1)$.

But $a_1$ has two eigenvalues (so $a_2$ also has two eigenvalues), thus $\Omega(A)=\sigma(a_1)$ has two elements, but $\sigma(a_1)\times\sigma(a_2)$ has 4 elements. This means precisely that the function $\Gamma$ you defined is not surjective in this case.

  • $\begingroup$ Thanks for your counterexample. $\endgroup$ – niki Dec 6 '14 at 18:39

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